Math 618 Theory of Functions of a Complex Variable II Spring 2015 Exercise on three-circles theorems In this assignment, you will apply subharmonic functions to prove Hadamard’s three-circles theorem (which is one of the topics on the official syllabus for the qualifying examination in complex analysis). 1. Suppose π’ is a subharmonic function in an annulus with inner radius π and outer radius π: namely, { π§ ∈ β βΆ π < |π§| < π }. Let π(π) denote the maximum of π’(π§) when |π§| = π. Prove that if π < π1 < π < π2 < π, then π(π) ≤ log π − log π1 log π2 − log π π(π1 ) + π(π2 ). log π2 − log π1 log π2 − log π1 Sometimes this inequality is described by saying that π(π) is “a convex function of log π.” An equivalent formulation of the inequality is that log π(π) ≤ log π2 π π2 π1 π(π1 ) + log ππ 1 log π2 π1 π(π2 ). Suggestion: Use a harmonic comparison function of the form π΄ + π΅ log |π§|. 2. Deduce that if π is a holomorphic function in the annulus, and π(π) denotes the maximum of |π (π§)| when |π§| = π, then π 1−π π(π) ≤ π(π1 ) π(π2 ) log , where π < π1 < π < π2 < π, and π = log π2 π π2 π1 . This statement is Hadamard’s three-circles theorem (Theorem 23.0.3 on page 387 of the textbook). 3. Suppose in Part 1 that π’ is bounded above, and π = 0. Let π1 approach 0 to deduce that π(π) is an increasing function of π. Extend π’ to the whole disk { π§ ∈ β βΆ |π§| < π } by declaring π’(0) to be limπ→0+ π(π). Prove that the extended function π’ is subharmonic on the whole disk. This statement is a removable singularity theorem for subharmonic functions that are bounded above on a punctured disk. April 2, 2015 Page 1 of 1 Dr. Boas